1
$\begingroup$

What really is a radical? For example, when $\ce{Cl2}$ receives enough energy to dissociate the bond and turn into $\ce{ 2Cl.}$, the two chlorine atoms each carry one of the electrons that form the bond. What is the difference between the $\ce{Cl}$ radical and the original $\ce{Cl}$? They have exactly the same number of electrons. And how do they act as a chain carrier?

$\endgroup$
4
$\begingroup$

free radical is a molecule with one or more unpaired electrons. Despite this definition, d-element compounds with unpaired electrons on d-element, like $\ce{CrCl3.6H2O}$ are not usually considered as free radicals because their unpaired electrons are relatively unreactive. Free radicals are especially reactive, because their reactions usually go through activated complex with relatively small energy. The reasons behind it are not that easy do describe.

In general most reaction with free radicals in organic chemistry falls into two categories: one, where free radical reacts with exposed hydrogen/heteroatom and one where it reacts with exposed multiple bond. In both cases, orbitals of interests are three orbitals of the radical in consideration and two atoms of the organic molecules. MO theory says, that such interaction results in three MOs: bonding, nonbonding and antibonding. With three electrons occupying this orbitals, the energy of the system is mostly same as for the radical and the molecule separated, so the activation energy of the reaction is near zero.

In case of nucleophilic/electrophilic reactions in , the interaction in activated complex are similar, with two electrons on bonding and zero or two electrons on nonbonding orbitals. However, in case of aliphatic derivatives, the reaction usually goes throw sterically hindered state with 5-coordinated carbon and/or requires generation of hi-energy particles that are not preserved after the molecule reacts. In cases when hi-energy intermediate may be preserved (for example, ionic polymerization), and/or reaction goes through sterically unhindered intermediates (for example, nucleophilic substitution in acyl halides), and/or one of the reagents easily produces active intermediates (for example, nitrating mixture in aromatic substitution) reactions without free radicals as intermediates may proceed swiftly.

Example of chain reaction for aliphatic radical substitution:

$\ce{Cl2 + CH4 -> CH3Cl + HCl}$

in case if $\ce{CH4}$ is presented in huge excess, resulting in reduced rate of multichlorination, the reaction proceeds as following:

  • Chain initiations: $\ce{Cl2 -> 2.Cl}$
  • Chain growth 1: $\ce{.Cl + CH4 -> HCl + .CH3}$
  • Chain growth g 2: $\ce{.CH3 + Cl2 -> CH3Cl + .Cl}$

In case of ionic polymerization

  • Chain initiation: $\ce{H2SO4 + CH2=CH-Ph -> HSO4- + H-CH_2-\stackrel{+}{C}H-Ph}$
  • Chain growth: $\ce{H-(CH2-CHPh-)_{n}-CH2-\stackrel{+}{C}H-Ph + CH2=CH-Ph ->} $ $\ce{H-(CH2-CHPh-)_{n + 1}-CH2-\overset{+}{C}H-Ph}$

Both reactions proceed smoothly at room temperature if source of active particles is available (UV-light for breaking $\ce{Cl-Cl}$ bond for the former and concentrated sulfuric acid for the latter)

Compare nucleophilic substitution $\ce{KOH + CH3I -> CH3OH + KI}$

  • $\ce{KOH -> K+ + OH-}$
  • $\ce{OH- + CH3-I -> (HO-CH3-I)- -> HO-CH3 + I-}$
  • $\ce{I- + K+ -> KI}$

As you see, the reaction proceeds through sterically hindered state with 5-coordinated carbon. Compare to acyl halide case $\ce{KOH + CH3COCl -> CH3COOH + KCl}$

  • $\ce{KOH -> K+ + OH-}$
  • $\ce{OH- + CH3-(C=O)-Cl -> CH3-C(O^{-})(OH)Cl -> CH3-COOH + Cl-}$
  • $\ce{Cl- + K+ -> KCl}$

in this case the reaction proceeds through sterically unhindered state with 4-coordinated carbon, so acetylchloride reacts with water at room temperature.

TL;DR : free radicals are particles with unpaired electron, readily available for covalent interactions, they reacts with molecules with easy because they are ones of big family of 'active particles' with low energy of activation of many reactions, and, unlike many other active species, they usually regenerate themselves in reaction, allowing chain reactions to proceed. Other active particles or plainly active substrates also exists, so free radicals are not special. The fact of presence of unpaired electron does not necessary make the particle an active free radical, many relatively inactive free radicals exists, like $\ce{NO}$.

Further reading: MO theory (case of 3-atomic interactions) and kinetics (case of chain reactions) parts of your favorite advance chemistry textbook, parts on reaction mechanisms and active particles of your favorite advanced organic chemistry textbook.

Despite above answer based exclusively on organic chemistry examples, everything stated above is true for inorganic chemistry. However the set of possible kinds of elementary acts, active particles and interactions is much wider than for basic organic chemistry, leading to many special and unique cases and consequently to much less predictable reactions in complicated cases.

$\endgroup$
1
$\begingroup$

A radical is a species with an unpaired electron. The chlorine molecule has a covalent bond joining the 2 chlorine atoms, there are no free or unpaired electrons. Because of this, the chlorine molecule is a stable species and energy must be put into the system to break the bond. If you break that bond you now have a chlorine atom with an unpaired electron. Such a radical with the unpaired electron is a very reactive species because of the unpaired electron. When such a radical reacts with a double bond a new free radical is generated at the other end of the double bond and chain propagation continues.

$\endgroup$
0
$\begingroup$

The difference between the Cl2 and the Cl (chlorine and chloride respectively) is that the in the chlorine MOLECULE the chlorine atoms are covalently bonded to each other thus filling all the occupied orbitals as one 3p electron from each Chlorine (2 in total) is shared between the two. As a result, one can take each chlorine atom's electron configuration to be 1s2, 2s2, 2p6, 3s2, 3p6. (identical to the unreactive noble gas Argon) When the covalent bond is broken, typically using UV light, it is broken by homolytic fission which means that one of the electrons from the two used in the covalent bond goes to each chlorine atom leaving them uncharged. However, if one takes a look at the electron configuration of the atom now - 1s2, 2s2, 2p6, 3s2, 3p5 - it's clear that the 3p orbital is incomplete (one p orbital is only occupied by one electron). This renders the Chlorine atom a Chlorine radical; there is no distinction between a chlorine atom and a chlorine radical. This also explains why chlorine is diatomic (is mostly present as a Cl2 molecule) with a single covalent bond. This is similar the same as all group 7 elements. Also similar to elements like oxygen which are di-radical as there are two unpaired electrons in the the 3p subshell. This explains why oxygen molecules have a double bond and are also diatomic. I know "two un**paired** electrons" sounds counter intuitive but the p sub-shells are filled in such a way as to avoid pairing the electrons in an orbital until absolutely necessasry (i.e once three electrons are already in the p sub-shell). N2 (Nitrogen) follows the pattern, forming triple bonds in a diatomic molecule to pair electrons in orbitals.

Radicals are extremely reactive as a result. They're chain carriers in reactions called radical substitution reactions (search "radical substitution chemguide" on google). An example is the production of halogenoalkanes from alkanes which are useful for synthesis of alcohols and amines.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.